<ruby id="bdb3f"></ruby>

    <p id="bdb3f"><cite id="bdb3f"></cite></p>

      <p id="bdb3f"><cite id="bdb3f"><th id="bdb3f"></th></cite></p><p id="bdb3f"></p>
        <p id="bdb3f"><cite id="bdb3f"></cite></p>

          <pre id="bdb3f"></pre>
          <pre id="bdb3f"><del id="bdb3f"><thead id="bdb3f"></thead></del></pre>

          <ruby id="bdb3f"><mark id="bdb3f"></mark></ruby><ruby id="bdb3f"></ruby>
          <pre id="bdb3f"><pre id="bdb3f"><mark id="bdb3f"></mark></pre></pre><output id="bdb3f"></output><p id="bdb3f"></p><p id="bdb3f"></p>

          <pre id="bdb3f"><del id="bdb3f"><progress id="bdb3f"></progress></del></pre>

                <ruby id="bdb3f"></ruby>

                企業??AI智能體構建引擎,智能編排和調試,一鍵部署,支持知識庫和私有化部署方案 廣告
                # 0349. 兩個數組的交集 ## 題目地址(349. 兩個數組的交集) <https://leetcode-cn.com/problems/intersection-of-two-arrays/> ## 題目描述 ``` <pre class="calibre18">``` 給定兩個數組,編寫一個函數來計算它們的交集。 示例 1: 輸入:nums1 = [1,2,2,1], nums2 = [2,2] 輸出:[2] 示例 2: 輸入:nums1 = [4,9,5], nums2 = [9,4,9,8,4] 輸出:[9,4] 說明: 輸出結果中的每個元素一定是唯一的。 我們可以不考慮輸出結果的順序。 ``` ``` ## 前置知識 - hashtable ## 公司 - 阿里 - 騰訊 - 百度 - 字節 ## 思路 先遍歷第一個數組,將其存到hashtable中, 然后遍歷第二個數組,如果在hashtable中存在就push到return,然后清空hashtable即可。 ## 關鍵點解析 - 空間換時間 ## 代碼 - 語言支持:JS, Python Javascript Code: ``` <pre class="calibre18">``` <span class="hljs-title">/** * @param {number[]} nums1 * @param {number[]} nums2 * @return {number[]} */</span> <span class="hljs-keyword">var</span> intersection = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">nums1, nums2</span>) </span>{ <span class="hljs-keyword">const</span> visited = {}; <span class="hljs-keyword">const</span> ret = []; <span class="hljs-keyword">for</span>(<span class="hljs-keyword">let</span> i = <span class="hljs-params">0</span>; i < nums1.length; i++) { <span class="hljs-keyword">const</span> num = nums1[i]; visited[num] = num; } <span class="hljs-keyword">for</span>(<span class="hljs-keyword">let</span> i = <span class="hljs-params">0</span>; i < nums2.length; i++) { <span class="hljs-keyword">const</span> num = nums2[i]; <span class="hljs-keyword">if</span> (visited[num] !== <span class="hljs-params">undefined</span>) { ret.push(num); visited[num] = <span class="hljs-params">undefined</span>; } } <span class="hljs-keyword">return</span> ret; }; ``` ``` Python Code: ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">intersection</span><span class="hljs-params">(self, nums1: List[int], nums2: List[int])</span> -> List[int]:</span> visited, result = {}, [] <span class="hljs-keyword">for</span> num <span class="hljs-keyword">in</span> nums1: visited[num] = num <span class="hljs-keyword">for</span> num <span class="hljs-keyword">in</span> nums2: <span class="hljs-keyword">if</span> num <span class="hljs-keyword">in</span> visited: result.append(num) visited.pop(num) <span class="hljs-keyword">return</span> result <span class="hljs-title"># 另一種解法:利用 Python 中的集合進行計算</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">intersection</span><span class="hljs-params">(self, nums1: List[int], nums2: List[int])</span> -> List[int]:</span> <span class="hljs-keyword">return</span> set(nums1) & set(nums2) ``` ``` **復雜度分析** - 時間復雜度:O(N)O(N)O(N) - 空間復雜度:O(N)O(N)O(N) 更多題解可以訪問我的LeetCode題解倉庫:<https://github.com/azl397985856/leetcode> 。 目前已經37K star啦。 關注公眾號力扣加加,努力用清晰直白的語言還原解題思路,并且有大量圖解,手把手教你識別套路,高效刷題。 ![](https://img.kancloud.cn/cf/0f/cf0fc0dd21e94b443dd8bca6cc15b34b_900x500.jpg)
                  <ruby id="bdb3f"></ruby>

                  <p id="bdb3f"><cite id="bdb3f"></cite></p>

                    <p id="bdb3f"><cite id="bdb3f"><th id="bdb3f"></th></cite></p><p id="bdb3f"></p>
                      <p id="bdb3f"><cite id="bdb3f"></cite></p>

                        <pre id="bdb3f"></pre>
                        <pre id="bdb3f"><del id="bdb3f"><thead id="bdb3f"></thead></del></pre>

                        <ruby id="bdb3f"><mark id="bdb3f"></mark></ruby><ruby id="bdb3f"></ruby>
                        <pre id="bdb3f"><pre id="bdb3f"><mark id="bdb3f"></mark></pre></pre><output id="bdb3f"></output><p id="bdb3f"></p><p id="bdb3f"></p>

                        <pre id="bdb3f"><del id="bdb3f"><progress id="bdb3f"></progress></del></pre>

                              <ruby id="bdb3f"></ruby>

                              哎呀哎呀视频在线观看